Transcript 1.3 – Continuity, End Behavior, and Limits

1.3 – Continuity, End Behavior,
and Limits
Ex. 1 Determine whether each function is
continuous at the given x value(s). Justify using
the continuity test. If discontinuous, identify the
type of discontinuity as infinite, jump, or
removable.
a. f(x) = 3x – 2 if x > -3 ; at x = -3
2 – x if x < - 3
Ex. 1 Determine whether each function is
continuous at the given x value(s). Justify using
the continuity test. If discontinuous, identify the
type of discontinuity as infinite, jump, or
removable.
a. f(x) = 3x – 2 if x > -3 ; at x = -3
2 – x if x < - 3
1. Find f(-3).
Ex. 1 Determine whether each function is
continuous at the given x value(s). Justify using
the continuity test. If discontinuous, identify the
type of discontinuity as infinite, jump, or
removable.
a. f(x) = 3x – 2 if x > -3 ; at x = -3
2 – x if x < - 3
1. Find f(-3).
f(-3) = 2 – (-3) = 5
Ex. 1 Determine whether each function is
continuous at the given x value(s). Justify using
the continuity test. If discontinuous, identify the
type of discontinuity as infinite, jump, or
removable.
a. f(x) = 3x – 2 if x > -3 ; at x = -3
2 – x if x < - 3
1. Find f(-3).
f(-3) = 2 – (-3) = 5, so f(-3) exists
2. Investigate values close to f(-3)
2. Investigate values close to f(-3)
x
-3.1
-3.01
-3.001
f(x)
5.1
5.01
5.001
-3
-2.999
-2.99
-2.9
-10.997
-10.97
-10.7
2. Investigate values close to f(-3)
x
-3.1
-3.01
-3.001
f(x)
5.1
5.01
5.001
As x
-3
-3 from left, f(x)
-2.999
-2.99
-2.9
-10.997
-10.97
-10.7
5
2. Investigate values close to f(-3)
x
-3.1
-3.01
-3.001
f(x)
5.1
5.01
5.001
As x
As x
-3
-2.999
-2.99
-2.9
-10.997
-10.97
-10.7
-3 from left, f(x) 5
-3 from right, f(x) -11
2. Investigate values close to f(-3)
x
-3.1
-3.01
-3.001
f(x)
5.1
5.01
5.001
-3
-2.999
-2.99
-2.9
-10.997
-10.97
-10.7
As x -3 from left, f(x) 5
As x -3 from right, f(x) -11
Since don’t approach same value,
discontinuous and jump discontinuity.
b. f(x) = x + 3
x2 – 9
; at x = -3 and x = 3
b. f(x) = x + 3
x2 – 9
; at x = -3 and x = 3
1. Find f(-3) and f(3).
b. f(x) = x + 3
x2 – 9
; at x = -3 and x = 3
1. Find f(-3) and f(3).
f(-3) = -3 + 3 = 0 = Ø
(-3)2 – 9 0
b. f(x) = x + 3
x2 – 9
; at x = -3 and x = 3
1. Find f(-3) and f(3).
f(-3) = -3 + 3 = 0 = Ø
(-3)2 – 9 0
f(3) = 3 + 3 = 6 = Ø
(3)2 – 9
0
b. f(x) = x + 3
x2 – 9
; at x = -3 and x = 3
1. Find f(-3) and f(3).
f(-3) = -3 + 3 = 0 = Ø
(-3)2 – 9 0
f(3) = 3 + 3 = 6 = Ø
(3)2 – 9
0
Since both f(-3) = Ø and f(3) = Ø, f(x) is
discontinuous at both x = -3 and x = 3.
2. Investigate values close to f(-3) and f(3).
2. Investigate values close to f(-3) and f(3).
x
f(x)
-3.1
-3.01
-3.001
-0.164 -0.166 -0.167
-3
-2.999
-2.99
-2.9
-0.167
-0.167 -0.169
2. Investigate values close to f(-3) and f(3).
x
f(x)
As x
-3.1
-3.01
-3.001
-3
-0.164 -0.166 -0.167
-3 from left, f(x)
-2.999
-2.99
-0.167
-0.167 -0.169
-0.167
-2.9
2. Investigate values close to f(-3) and f(3).
x
f(x)
As x
As x
-3.1
-3.01
-3.001
-0.164 -0.166 -0.167
-3
-2.999
-2.99
-0.167
-0.167 -0.169
-3 from left, f(x) -0.167
-3 from right, f(x) -0.167
-2.9
2. Investigate values close to f(-3) and f(3).
x
f(x)
-3.1
-3.01
-3.001
-0.164 -0.166 -0.167
-3
-2.999
-2.99
-2.9
-0.167
-0.167 -0.169
As x -3 from left, f(x) -0.167
As x -3 from right, f(x) -0.167
Since they approach same value, limit exists.
2. Investigate values close to f(-3) and f(3).
x
f(x)
-3.1
-3.01
-3.001
-3
-0.164 -0.166 -0.167
-2.999
-2.99
-2.9
-0.167
-0.167 -0.169
As x -3 from left, f(x) -0.167
As x -3 from right, f(x) -0.167
Since they approach same value, limit exists.
x
2.9
2.99
2.999
f(x)
-10
-100
-1000
3.0
3.001
3.01
3.1
1000
100
10
2. Investigate values close to f(-3) and f(3).
x
f(x)
-3.1
-3.01
-3.001
-3
-0.164 -0.166 -0.167
-2.999
-2.99
-2.9
-0.167
-0.167 -0.169
As x -3 from left, f(x) -0.167
As x -3 from right, f(x) -0.167
Since they approach same value, limit exists.
x
2.9
2.99
2.999
f(x)
-10
-100
-1000
As x
3.0
3 from left, f(x)
3.001
3.01
3.1
1000
100
10
-∞
2. Investigate values close to f(-3) and f(3).
x
f(x)
-3.1
-3.01
-3.001
-3
-0.164 -0.166 -0.167
-2.999
-2.99
-2.9
-0.167
-0.167 -0.169
As x -3 from left, f(x) -0.167
As x -3 from right, f(x) -0.167
Since they approach same value, limit exists.
x
2.9
2.99
2.999
f(x)
-10
-100
-1000
As x
As x
3.0
3 from left, f(x)
3 from right, f(x)
3.001
3.01
3.1
1000
100
10
-∞
∞
2. Investigate values close to f(-3) and f(3).
x
f(x)
-3.1
-3.01
-3.001
-3
-0.164 -0.166 -0.167
-2.999
-2.99
-2.9
-0.167
-0.167 -0.169
As x -3 from left, f(x) -0.167
As x -3 from right, f(x) -0.167
Since they approach same value, limit exists.
x
2.9
2.99
2.999
f(x)
-10
-100
-1000
3.0
3.001
3.01
3.1
1000
100
10
As x 3 from left, f(x)
-∞
As x 3 from right, f(x)
∞
Since limit x -3 exists but f(-3) doesn’t,
removable discontinuity.
2. Investigate values close to f(-3) and f(3).
x
f(x)
-3.1
-3.01
-3.001
-3
-0.164 -0.166 -0.167
-2.999
-2.99
-2.9
-0.167
-0.167 -0.169
As x -3 from left, f(x) -0.167
As x -3 from right, f(x) -0.167
Since they approach same value, limit exists.
x
2.9
2.99
2.999
f(x)
-10
-100
-1000
3.0
3.001
3.01
3.1
1000
100
10
As x 3 from left, f(x)
-∞
As x 3 from right, f(x)
∞
Since limit x -3 exists but f(-3) doesn’t,
removable discontinuity.
Since limit x -3 doesn’t exist, infinite
discontinuity.
Ex. 2 Use the graph of the function to describe
its end behavior.
Ex. 2 Use the graph of the function to describe its
end behavior.
lim f(x) = - ∞
x -∞
Ex. 2 Use the graph of the function to describe its
end behavior.
lim f(x) = - ∞
x -∞
lim f(x) = - ∞
x -∞